Let $a, b, c$ - the positive real numbers, and $ab+bc+ca=1$ Prove that $\sqrt{a+\frac{1}{a}}+\sqrt{b+\frac{1}{b}}+\sqrt{c+\frac{1}{c}} \geqslant 2(\sqrt{a}+\sqrt{b}+\sqrt{c})$
Probably, we should use these facts:
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = \frac{1}{abc}$
$(a+b+c)^2 = a^2 + b^2 + c^2 + 2$
But I don't know how to use them. Please, help.
We have $$ \sum_{cyc} \sqrt{a+\frac{1}{a}}=\sum_{cyc} \sqrt{a+\frac{ab+bc+ca}{a}}=\sum_{cyc} \sqrt{a+\frac{bc}{a}+b+c}\ge\sum_{cyc} \sqrt{2\sqrt{bc}+b+c}=\sum_{cyc} \sqrt{\left(\sqrt b+\sqrt c\right)^2}=\sum_{cyc} \left(\sqrt b+\sqrt c\right)=2\left(\sqrt a+\sqrt b+\sqrt c\right) $$ as desired.
Note: The inequality is due to the well known inequality between the arithmetic mean and the geometric mean, which states that: $$ \frac{a_1+…+a_n}{n}\ge\left(a_1\cdots a_n\right)^\frac{1}{n} $$ Taking $n=2$, $a_1=a$ and $a_2=\frac{bc}{a}$ this yields: $$ \frac{a+\frac{bc}{a}}{2}\ge\left(a\cdot\frac{bc}{a}\right)^\frac12=\sqrt{bc}\iff a+\frac{bc}{a}\ge2\sqrt{bc} $$ This is, as shown above, enough to prove your inequality.
By Cauchy-Schwarz we have $$ \sqrt{a + \frac{1}{a}} = \sqrt{a + b + c + \frac{bc}{a}} \geq \frac{1}{2}\left(\sqrt{a} + \sqrt{b} + \sqrt{b} + \sqrt{\frac{bc}{a}}\right) $$ and similarly $$ \sqrt{b + \frac{1}{b}} \geq \frac{1}{2}\left(\sqrt{a} + \sqrt{b} + \sqrt{b} + \sqrt{\frac{ca}{b}}\right) $$ and $$ \sqrt{c + \frac{1}{c}} \geq \frac{1}{2}\left(\sqrt{a} + \sqrt{b} + \sqrt{b} + \sqrt{\frac{ab}{c}}\right). $$ Thus $$ \sqrt{a + \frac{1}{a}} + \sqrt{b + \frac{1}{b}} + \sqrt{c + \frac{1}{c}} \geq \frac{1}{2}\left(3\left(\sqrt{a} + \sqrt{b} + \sqrt{c}\right) + \frac{\sqrt{a}\sqrt{b}}{\sqrt{c}} + \frac{\sqrt{b}\sqrt{c}}{\sqrt{a}} + \frac{\sqrt{c}\sqrt{a}}{\sqrt{b}} \right) \geq 2\left(\sqrt{a} + \sqrt{b} + \sqrt{c}\right). $$
